Hypothesis Testing One Sample

8/11/2019 Hypothesis Testing One Sample

1/78

HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1

PREVIOUSLY

used confidence intervals to answer questions such as...

You know that 0.25% of women have red/green color

blindness. You conduct a study of men and find that of 80 men

tested, 7 have red/green color blindness.

Based on the above information, do men have a higher

percentage of red/green color blindness than women? Why or

why not?

question is whether a sample proportion differs from a

population proportion


2/78


You measure body mass index (BMI) for 25 men and 25 women

and calculate the following statistics...

gender mean standard deviationwomen 25 6

men 26 3

The upper-limit for a NORMAL BMI for adults is 23. Based on

the above information, can you say that either the group of

women or group of men have 'above normal' BMIs? Why or

why not?

question is whether sample means differ from a population

mean


3/78


X ME

X t s

n

women

men

= =

= =

( )

: ( . ) . ( . , . )

: ( .)

. ( . , . )

,.24 025

25 2 064 625

25 2 477 22 523 27 477

26 2 064 3

2526 1238 24 762 27 238

confidence interval approach ...

women's 95% confidence interval includes 23 ... not different

men's 95% confidence interval above 23 ... above normal


4/78


You conduct health exams on samples of 25 men and 25

women. One measurement you make is systolic blood pressure

(SBP) and you calculate the following statistics...

gender mean standard deviation

women 120 15men 130 10

Do you think that men have a higher mean SBP than women?

Why or why not?

question is whether sample means differ from each other


5/78


X ME

X t s

n

women

men

= =

= =

( )

: ( . ) . ( . , . )

: ( .

)

. ( . , . )

,.24 025

120 2 064 15

25120 6192 113808 126192

130 2 064 10

25

130 4 128 125872 134 128

confidence interval approach ...

confidence intervals overlap ... not different


6/78


ANOTHER APPROACH TO SUCH QUESTIONS

hypothesis testing (almost always results in the same answer as

confidence intervals - exception possible with proportions dueto difference in how standard errors are calculated in hypothesis

testing versus confidence intervals)

one sample hypothesis testing

does a sample value differ from a population value

two sample hypothesis testing

do two sample values differ from each other


7/78


DEFINITIONS

Triola hypothesisis a claim or statement about a property

of a population

Daniel hypothesisis a statement about one or more

populations

Rosner no explicit definition, but ... preconceived ideas as towhat population parameters might be ...


8/78


Triola hypothesis test is a standard procedure for testinga claim about a property of a population

Daniel no definition

Rosner no explicit definition, but ... hypothesis testing

provides an objective framework for making decisions

using probabilistic methods rather than relying on

subjective impressions


9/78


COMPONENTS OF HYPOTHESIS TEST

... a problem with CONTINUOUS data ...

same problem ... you measure body mass index (BMI) for 25

men and 25 women and calculate the following statistics ...

gender mean standard deviationwomen 25 6

men 26 3

The upper-limit for a NORMAL BMI for adults is 23. Based on

the above information, can you say that either the group of

women or group of men have 'above normal' BMIs? Why or

why not?


10/78


start with women

given a claim ... women have above normal BMIs (above 23)

identify the null hypothesis ... the BMI of women is equal to 23

identify the alternative hypothesis ... the BMI of women isgreater than 23

express the null and alternative hypothesis is symbolic form ...

H0: = 23

H1: > 23


11/78


given a claim and sample data, calculate the value of the teststatistic ...

t

X

S N=

=

=

/ / .

25 23

6 25 167

given a significance level, identify the critical value(s)...

t-distribution with 24 degrees of freedom, , 1-tail test=.05

from table 5 in Rosner, critical value = 1.711

critical value method... test statistic < critical value

conclusion ... no evidence to say women are greater than

the population value of 23


12/78


1 711 0 05 1 67 1 318 0 10. ( . ) . . ( . ) = > > =

given a value of the test statistic, identify the P-value...

in the t-distribution, what is the probability of obtaining the value of the test

statistic (1.67) with 24 degrees of freedom

interpolate in table 5 in Rosner, look on the linewith 24 degrees of freedom and see that 1.67 lies

between 0.90 and 0.95 for the area in one-tail,

between the values ...

or use STATCRUNCH to get an exact value ...

P = 0.054

P-value method ...P > 0.05

conclusion ... no evidence to say women are greater thanthe population value of 23


13/78


state the conclusion of the hypothesis is simple, non-technicalterms ...

there is no evidence to conclude that the BMI of women is not

equal to 23

identify the type I and type II errors that can be made when

testing a given claim...

type I error determined by choice of (in this case, 0.05)

type II error is a function of...

sample size

variability (standard deviation)

difference worth detecting

size of chosen level of type I error


14/78


calculate type II error (formulas, SAS, web, etc.)

from a web site ... on class web

site, the link is ...

Rollin Brant's Sample Size/Power

Calculators (University of

Calgary)

or Rosner equation 7.19 ...

( / ) ( . / )

( . . ) ( . )

+ = +

+ =

z n

1 0 1 1645 23 25 25 6

1645 1667 0 021

from table 5 in Rosner, column A when z (or x in table) = 0.02 is 0.508

approximately a 50% chance of detecting a difference of 2 given the sample

size, variability, and type I error (type II error = $= 1-power = 0.492)

more about POWER and Type II error later ...


15/78

HYPOTHESIS TESTING (ONE SAMPLE) CHAPTER 7 16


16/78



in the t-distribution, what is the probability of obtaining the value of the test

statistic (5) with 24 degrees of freedom

in table 5 in Rosner, look on the line with 24degrees of freedom and see that 5 is to the right of

the value (3.745) for "=0.0005 (0.9995 in thetable) ... the P-value is less than 0.0005

or use STATCRUNCH to get an exact value ...

P = 2.0784282E-5

P-value method ...P < 0.05

conclusion ... men have a BMI greater than the population

value of 23



17/78



the BMI of men is greater than 23





sample sizevariability (standard deviation)





18/78





Rollin Brant's Sample Size/PowerCalculators (University of Calgary)


( / ) ( . / )

( . ) ( . )

+ = +

+ =

z n

1 0 1 1645 23 26 25 3

1645 5 3 355

from table 5 in Rosner, column A when z (or x in table) = 3.36 is 0.996

a 100% chance of detecting a difference of 3 given the sample size,

variability, and type I error (type II error = $= 1-power = 0.004)




19/78


using Statcrunch ... confidence interval and hypothesis test

women ...

men ...



20/78


confidence interval approach should use a one-sidedrather than a two-sided interval ... it should be an upper

one-sided interval

X ME

X t s

n

24 05,.

women... . . ( . , )25 1711 6

25 25 2 053 22 947

= =

men... . . ( . , )26 1711 3

2526 1 027 24 973

= =

women's 95% confidence interval includes 23 ... not different

men's 95% confidence interval above 23 ... above normal

still the same conclusion as critical and P-value methods



21/78


... a problem with DISCRETE data ...

Mendel's genetics experiment ... given 580 offspring peas and 26.2%

(N=152) with yellow pods ... what should one conclude about Mendel'stheory that 25% of peas will have yellow pods

NPQ = 580(0.25)(0.75) = 109

(note: previously, check of NPQ used SAMPLE PROPORTION ... check of NPQ

now uses POPULATION PROPORTION since population proportion is assumed

to be known ... discussed on bottom of page 205 in Rosner)

express the null and alternative hypothesis is symbolic form ...

H0: p = 0.25

H1: p 0.25



22/78


z

p p

p q n=

=

=

$

/

. .

( . )( . ) / .0

0 0

0 262 0 25

0 25 0 75 580 0 67

given a claim and sample data, calculate the value of the test statistic ...

given a significance level, identify the critical value(s)...

normal distribution (z), , 2-tail test=.05from table 5 in Rosner, critical value = 1.96

critical value method... test statistic < critical value

conclusion ... there is no evidence to conclude that the proportion of

yellow pods is different from 0.25



23/78



in the normal (z) distribution, what is the probability of obtaining the value of

the test statistic (0.67)

use table 3 in Rosner, find the value of the test

statistic in column x, then look in column B and

see that the area in the tail distribution is

0.2514

you can use STATCRUNCH to get the figure on

the right

P-value method ...P > 0.05

conclusion ... there is no evidence to conclude that the proportion of

yellow pods is different from 0.25



24/78



there is no evidence to conclude that the proportion of yellow

peas is different from 25%





sample size

variability (standard deviation)





25/78





Rollin Brant's Sample Size/PowerCalculators (University of Calgary)


p q

p qz

p p n

p q

0 0

1 1

20 1

0 0

25 75

262 738196

25 262 580

25 751248

/

| | (. )(. )

(. )(. ).

|. . |

(. )(. )( . )+

= +

=

from table 5 in Rosner, column B when z (or x in table) = 1.25 is 0.1056

a 10% chance of detecting a difference of 0.012 given the sample size,

variability, and type I error (type II error = $= 1-power = 0.8944)




26/78


there is no evidence to conclude that the proportion of yellow

pods is different from 0.25

why is the power so low?

to detect a difference of 0.012 with power = 0.80, n=10,316



27/78


confidence interval approach ... two-sided interval since the

alternative hypothesis is two-sided

NPQ = 580(0.262)(0.738) = 112

(note: check of NPQ uses SAMPLE PROPORTION since population proportionis assumed to be unknown ... discussed on bottom of page 205 in Rosner)

standard error =$ $ . ( . )

.pq

n

= =0262 0738

580

0 01826

95% confidence interval, Z = 1.96

margin of error, E = 1.96 * 0.01826 = 0.03579

P - E = 0.262 - 0.03579 = 0.226

P+ E = 0.262 + 0.03570 = 0.298

95% confidence interval ... 0.226 < P < 0.298

conclusion ... no evidence to say that the population proportion is

different from 0.25 since it is within the confidenceinterval



28/78

( )

COMPONENTS OF HYPOTHESIS TESTING

before ...

what are the null and alternative hypotheses (what is your claim

about your data)

what is the appropriate test statistic

what is (are) the critical value(s) (type I error, type II error,power)

after ...

what is the P-value (what is/are the confidence interval(s))

what is the decision (conclusion) ... critical value, P-value

(confidence interval)



29/78

( )

null and alternative hypotheses

the null hypothesis for a one sample test ALWAYS contains a statement that

the value of a population parameter is EQUAL to some claimed value

the alternative hypothesis can contain a statement that a the value of a

population is either NOT EQUAL to, LESS than, or GREATER than some

claimed value (two-tailed and one-tailed hypothesis tests)

implication of above on your own claim ... your claim might be the null or it

might be the alternative hypothesis ... a product called "Gender Choice"

increases the probability of a female infant, or ... P(female) > 0.50

H0: p = 0.50

H1: p > 0.50



30/78

( )

test statistic

proportion z p p p q n= ( $ ) / /0 0 0

ALWAYS Z (note: POPULATION PROPORTIONS)

mean z x n= ( ) / ( / )

t x s n= ( ) / ( / )

Z OR t DEPENDS ON WHETHER POPULATION

VARIANCE IS KNOWN (Z), UNKNOWN (t)

variance 2 2 21= ( ) /n

ALWAYS CHI-SQUARE



31/78

critical values

any value that separates the critical region (where the null hypothesis

is rejected) from values that do not lead to rejection of the null

hypothesis

depends on alternative hypothesis and selected level of type I error

not equal - two-tail test, 50% of area in each tail of the distribution

greater or less than - one-tail test, 100% of area in one tail of the

distribution

two-tail test, test statistic is z, , critical value 1.96=0 05.

one-tail test, test statistic is z, , critical value 1.645

=0 05.



32/78

P-value

the probability of obtaining a value of the test statistic that is at

least as extreme as the one calculated using sample data given

that the null hypothesis is true

calculation of an exact P-value depends on the alternativehypothesis

not equal - two-tail test, double value found in appropriate table

greater or less than - one-tail test, value found in appropriate

table



33/78

figure from Triola ... how to determine P-values



34/78

decisions and conclusions

Critical Value Method

reject H0if the test statistic falls within the critical region, fail to

reject H0if the test statistic does not fall within the critical

region

P-Value Method

reject H0if P-value

fail to reject H0if P-value >

(or report the P-value and leave the decision to the reader)



35/78

Confidence Interval

if the confidence interval contains the likely value of the

population parameter, reject the claim that the population

parameter has a value that is not included in the confidence

interval

Rosner - if testing H0: versus the alternative hypothesis = 0H1: , H0is rejected if and only if a two-sided confidence 0interval for does not contain 0

Rosner - significance levels and confidence limits provide

complimentary information and both should be reported where

possible



36/78

figure from Triola ... wording a conclusion based on a

hypothesis test



37/78

two types of error

TYPE I AND TYPE II ERRORS

type I error - reject the null when it is actually true ( )

type II error - fail to reject the null when it is actually false ( )

power = 1 - (the probability of rejecting a false H0)



38/78

a look at power...



39/78

how does this diagram relate to the question about women's BMI...

the UPPER DIAGRAM --- given a hypothesized mean of 23, a standard

deviation of 6, a sample size of 25, and a one-tail test withalpha=0.05, any value greater than 24.974 would be called

"significantly greater" than 23 (would fall in the critical region)...

( . ) / ( / ) .24 974 23 6 25 1 645 =

consider 24.974 ~ 25, therefore, you would not reject the null

hypothesis given any sample mean less than 25

the LOWER DIAGRAM --- if "reality" was a mean of 25, what portion of

the area of a standard normal distribution would lie to the right of 25

(or 24.974), answer is 50%, so the power ~ 0.50



40/78

another look at the same problem ... one-tail test, "=0.05

curve on left represents null

hypothesis ... :=23

curve on right represents :=25

dark shaded area represents 5% of

the area of the left curve, or

z=1.645 ... area where you reject

H0: :=23

lighter shaded area represents

approximately 51% of the area of

the right curve

therefore ... if "reality" is really the curve on the right, :=25, and youconduct an experiment with N=25, F=6, and "=0.05 ... you only have a

51% chance of rejecting the null, your hypothesized "reality" of :=23


41/78



42/78

formula from Rosner - given a two-sided test, sample size for a given

"and $...

n z z= + 2

1 1 22

1 02( ) / ( )/

problem 7.9 from Rosner... plasma glucose level among sedentary

people (check for diabetes)... is their level higher or lower than the

general population

among 35-44 year olds, = 4.86 (mg/dL), F= 0.54

if a difference of 0.10 is worth detecting, with "=0.05, what sample

size is needed to have 80% power ($=0.20)

n z z= +0 54 0 102

0 80 0 9752 2

. ( ) / ( . ). .

n= + =29 16 0 84 196 228 6 2292. ( . . ) . ~



43/78

from SAS...

Distribution Normal

Method Exact

Number of Sides 2

Null Mean 4.96

Alpha 0.05

Mean 4.86Standard Deviation 0.54

Nominal Power 0.8

Computed N TotalActual N

Power Total

0.800 231



44/78

from a web site ...



45/78

same data, different problem...

problem 7.9 from Rosner... plasma glucose level among

sedentary people (check for diabetes)... is their level higher or

lower than the general population

among 35-44 year olds, = 4.86 (mg/dL), F= 0.54

if a difference of 0.10 is worth detecting, with "=0.05, with asample size of 100, what is the power



46/78

( . . ) / ( . / ) .4 966 4 86 0 54 100 1 96 =

( . . ) / ( . / ) .4 754 4 86 0 54 100 1 96 =

same approach as the BMI problem...

given a hypothesized mean of 4.86, a standard deviation of

0.54, a sample size of 100, and a two-tail test with alpha=0.05,

any value greater than 4.966 (or less than 4.754) would be

called "significantly different" from 4.86 (would fall in the critical

region)...

if "reality" was a mean of 4.96, what portion of the area of astandard normal distribution would lie to the right of 4.966,

answer is ~46%, so the power ~0.46



47/78

another look at the same problem illustration ... shaded area on

the right is power ... you find a value greater than 4.86 ...


48/78



49/78

WHAT SHOULD YOU KNOW

NOT formulas !!!

what is power (from Rosner ... the probability of detecting asignificant difference ... or the probability of rejecting H0when

H1is true)

how to illustrate the concept of power (two normal curvesdrawn using information about the null hypothesis, ", samplesize, F, difference worth detecting)

what are Type I and Type II error

the relationships among ", $, n, F, | | 0



50/78

PROPORTION

all the previous comments about the importance of proper

sampling (random, representative)

all conditions for a binomial distribution are met (fixed n ofindependent trials with constant p and only two possible

outcomes for each trial)

NPQ $5 (note: P and Q are POPULATION values)



51/78

test statistics is ALWAYS z...

z p p p q n= $ / /0 0 0

example 7.26 from Rosner study guide...

area of current interest in cancer epidemiology is the possible role oforal contraceptives (OCs) in the development of breast cancer ... in a

group of 1000 premenopausal women ages 4049 who are current

users of OCs, 15 subsequently develop breast cancer over the next 5

years ... if the expected 5-year incidence rate of breast cancer in thisgroup is 1.2% based on national incidence rates, then test the

hypothesis that there is an association between current OC use and

the subsequent development of breast cancer



52/78

claim ... there is an association between OC use and subsequentdevelopment of breast cancer

NPQ ... 1000(.012)(.988) = 11.856 > 5

null and alternative hypothesis ...

H p0 0012: .=H p1 0012: .

calculate the test statistic ... note that POPULATION values are

used in the denominator ...

z p p p q n= = ( $ ) / / ( . . ) / ( . )( . ) /0 0 0 0 015 0 012 0 012 0 988 1000

z= =0 033 0 03344 0 87. / . .



53/78

given a significance level, identify the critical value...

"=0.05, z = 1.96 (two-tail test)

critical value method ... test statistic (0.87) < critical value

identify the P-value...

from table 3 in Rosner ... what is area in tails of normal curve

with z=0.87, P=0.1992 (from column B)

from figure page 271 in Rosner7.6, two-sided test, double the P-value, P=0.38 (> 0.05)

conclusion ... no evidence to reject the null hypothesis



54/78

what is the confidence interval ... note that SAMPLE VALUES are

used to compute the confidence interval ...

CI z pq n= = ( $ $/ ) . . ( . )( . ) /0 015 1 96 0 015 0 985 1000

CI= 0015. 0.0038

(0.011, 0.019)

same conclusion since null hypothesis value of 0.012 is withinthe confidence interval



55/78

from Triola ... a problem where a hypothesis test and confidenceintervals give different results

given ... n=1000, k=119, and null

value of p=0.10 ...

H p0

0 10: .=H p

1 010: .

hypothesis test, z > critical value of

1.96 and P-value < 0.05

however, confidence interval includesthe null value of P = 0.10

critical value method and P-value results not the same as the

confidence interval results ... why?



56/78

can also calculate theexact P-value using

the binomial

distribution ... this is

the equivalent of a

hypothesis test with

an alternative

hypothesis of ...

, notH p1

0 10: .>H p

1 0 10: .

therefore ...

P-value=2(P)=0.0556

what is the 'exact' conclusion?



57/78

MEAN: KNOWN



the value of the population standard deviation is known

the population is normally distributed or n > 30 (Central Limit

Theorem)


58/78



59/78

claim... the serum cholesterol of recent Asian immigrants differs

from that in the US population

null and alternative hypotheses...

H

H

0

1

190

190

:

:

=

calculate the test statistic...

z= = = ( . ) / ( / ) . / . .18152 190 40 200 8 48 2 828 3 00



60/78

given a significance level, identify the critical value...

"=0.05, z=1.96 (two-tail test)

critical value method ... test statistic (-3) less that critical value (-1.96)

identify the P-value...

from table 3 ...what is area in tails of normal curve with z=2.12, P=0.0170

two-sided test, double the P-value, P=0.034

conclusion...

reject the null hypothesis



61/78

what is the confidence interval ...

CI x z n= = = ( / ) . . ( / ) . . 18152 1 96 40 200 18152 5 54

(175.98, 187.06)

same conclusion since 190 is not within the confidence interval



62/78

MEAN: UNKNOWN



the value of the population standard deviation is unknown

the population is normally distributed or n > 30 (Central Limit

Theorem)



63/78

test statistic WITH FUNKNOWN...

t x s n= ( ) / ( / )

example 7.3 from Rosner study guide ...

as part of a dietary-instruction program, eight 2534-year-old

females report an average daily intake of saturated fat of 11 gwith standard deviation of 11 g while on a vegetarian diet ... if

the average daily intake of saturated fat among 2534-year-old

females in the general population is 24 g, then, using asignificance level of .01, test the hypothesis that the intake of

saturated fat in this group is lower than that in the general

population



64/78

claim ... the daily intake of saturated fat of a group of females

on a vegetarian diet is lower than that among females in the

general population

null and alternative hypotheses...

HH

0

1

2424

::

=

Date post:	02-Jun-2018
Category:	Documents
Upload:	anastasia
View:	225 times
Download:	0 times

Hypothesis Testing One Sample

Documents